## Exercises - Comparisons of Functions (and Review)

1. Evaluate the following limits; provide a graphical interpretation for each.

$$\textrm{(a) } \lim_{x \rightarrow -4} \frac{x}{(x+4)^2} \quad \textrm{(b) } \lim_{x \rightarrow -4} \frac{x^2}{(x+4)^2} \quad \textrm{(c) } \lim_{x \rightarrow -\infty} \frac{x}{(x+4)^2} \quad \textrm{(d) } \lim_{x \rightarrow -\infty} \frac{x^2}{(x+4)^2}$$

(a) $-\infty$, vertical asymptote at $x=-4$
(b) $+\infty$; vertical asymptote at $x=-4$
(c) $0$; horizontal asymptote at $y=0$
(d) $1$; horizontal asymptote at $y=1$

2. Evaluate $\displaystyle{\lim_{x \rightarrow \infty} \frac{x^2 - 3x +5}{7x^2 + 8x - 9}}$ and provide a graphical interpretation.

$$\lim_{x \rightarrow \infty} \frac{x^2 - 3x +5}{7x^2 + 8x - 9} = \lim_{x \rightarrow \infty} \frac{x^2 - 3x +5}{7x^2 + 8x - 9} \cdot \frac{\frac{1}{x^2}}{\frac{1}{x^2}} = \lim_{x \rightarrow \infty} \frac{1 - \frac{3}{x} + \frac{5}{x^2}}{7 + \frac{8}{x} - \frac{9}{x^2}} = \frac{1 - 0 + 0}{7 + 0 - 0} = \frac{1}{7}$$

Graphically, there is a horizontal asymptote (on the right) of $y = 1/7$.

3. Evaluate $\displaystyle{\lim_{x \rightarrow -\infty} \frac{(2x-3)^2}{5x^3 + 6x + 7}}$ and provide a graphical interpretation.

$$\lim_{x \rightarrow -\infty} \frac{(2x-3)^2}{5x^3 + 6x + 7} = \lim_{x \rightarrow -\infty} \frac{4x^2-12x+9}{5x^3 + 6x + 7} \cdot \frac{\frac{1}{x^3}}{\frac{1}{x^3}} = \lim_{x \rightarrow -\infty} \frac{\frac{4}{x}-\frac{12}{x^2}+\frac{9}{x^3}}{5 + \frac{6}{x^2} + \frac{7}{x^3}} = \frac{0 -0+0}{5+0+0} = 0$$

Graphically, there is a horizontal asymptote (on the left) of $y=0$.

4. Evaluate $\displaystyle{\lim_{x \rightarrow \infty} \left( \frac{x^2}{2x+1} \right)^2}$ and provide a graphical interpretation.

$$\lim_{x \rightarrow \infty} \left( \frac{x^2}{2x+1} \right)^2 = \lim_{x \rightarrow \infty} \left( \frac{x^2}{2x+1} \cdot \frac{\frac{1}{x}}{\frac{1}{x}}\right)^2 = \lim_{x \rightarrow \infty} \left( \frac{x}{2 + \frac{1}{x}} \right)^2 = +\infty$$

Graphically, the associated function grows in value without bound on the far right side of the graph.

5. Evaluate $\displaystyle{\lim_{x \rightarrow +\infty} \frac{\sqrt{x^6-2x^2+1}}{x^3+1}}$ and provide a graphical interpretation.

Bring the denominator inside the radical (which requires squaring it) and then divide the numerator and denominator of the resulting fraction inside the radical by the highest power of $x$ in the denominator (i.e., $x^6$).

This reveals a limiting value of $1$, suggesting there is a horizontal asymptote (on the right) of $y=1$.

6. Evaluate $\displaystyle{\lim_{x \rightarrow -\infty} \frac{\sqrt{x^6-2x^2+1}}{x^3+1}}$ and provide a graphical interpretation.

Bring the denominator inside the radical (which requires squaring it, and adding a factor of $-1$ outside the radical since $x \rightarrow -\infty$ allows us to assume $x$ is negative). Then divide the numerator and denominator of the resulting fraction inside the radical by the highest power of $x$ in the denominator (i.e., $x^6$).

This reveals a limiting value of $-1$, suggesting there is a horizontal asymptote (on the left) of $y=-1$.

7. Evaluate $\displaystyle{\lim_{x \rightarrow -3} \frac{|x^2 - 9|}{x^2+2x-3}}$ and provide a graphical interpretation.

Consider the left and right-sided limits separately. As these disagree, this limit does not exist. There is a gap discontinuity from $(-3,3/2)$ to $(-3,-3/2)$.

8. Evaluate $\displaystyle{\lim_{\theta \rightarrow \pi} \frac{1 + \cos \theta}{\sin \theta}}$ and provide a graphical interpretation.

Multiply the numerator and denominator by $(1-\cos x)$. This creates a $1 - \cos^2 x = \sin^2 x$ in the numerator which then allows for a cancellation of a factor of $\sin x$. The remaining limit can be found by direct substitution and equals $0$. Graphically, there is a hole at $(\pi,0)$.

9. Evaluate $\displaystyle{\lim_{\theta \rightarrow \pi/4} \frac{\sin \theta - \cos \theta}{1 - \tan \theta}}$ and provide a graphical interpretation.

Multiply numerator and denominator by $\cos x$ so the denominator is expressible as $\cos \theta - \sin \theta$ and cancels with the $\sin \theta - \cos \theta$ factor in the numerator, leaving a factor of $-1$.

This leaves $\displaystyle{\lim_{\theta \rightarrow \pi/4} -\cos \theta = -\frac{\sqrt{2}}{2}}$. Graphically, there is a hole at $(\pi/4,-\sqrt{2}/2)$.

10. Evaluate $\displaystyle{\lim_{x \rightarrow 1} \frac{\sqrt{2x+7} - 3}{x^2 - 1}}$ and provide a graphical interpretation.

Multiply the numerator and denominator by the conjugate $\sqrt{2x+7}+3$. This makes the numerator expressible as $2(x-1)$. Factoring the denominator reveals that one can cancel a common $(x-1)$ from the top and bottom, leaving an expression whose limit can be found by direct substitution.

The limiting value is $1/6$. Graphically, there is a hole at $(1,1/6)$.

11. Find the value of the limit $\displaystyle{\lim_{x \rightarrow 4} \frac{4\sqrt{x} - 8}{x-4}}$ and state its graphical interpretation.

$1$, hole at $(4,1)$

12. Find the value of the limit $\displaystyle{\lim_{x \rightarrow \frac{3\pi}{2}} \frac{\cos x}{1 + \sin x}}$ and state its graphical interpretation.

The limit does not exist (the left and right limits are infinite and different signs). Graphically, there is a vertical asymptote at $x =3\pi/2$.

13. Find the value of the limit $\displaystyle{\lim_{x \rightarrow -\infty} \frac{\sqrt{x^2 + 1}}{3x-4}}$ and state its graphical interpretation.

Bring the denominator inside the radical (which requires squaring it, and adding a factor of $-1$ outside the radical since $x \rightarrow -\infty$ allows us to assume $x$ is negative). Then divide the numerator and denominator of the resulting fraction inside the radical by the highest power of $x$ in the denominator (i.e., $x^2$).

This reveals a limiting value of $-1/3$ and graphically, a horizontal asymptote (on the left) of $y=-1/3$.

14. If $\displaystyle{\frac{10e^x - 21}{2e^x} \lt f(x) \lt \frac{5\sqrt{x}}{\sqrt{x-1}}}$ for all $x \gt 1$, compute $\displaystyle{\lim_{x \rightarrow \infty} f(x)}$.

Suspecting an application of the Squeeze Theorem, let us find the limiting values of the expressions on the left and right of this string inequality.

To do this on the left, one will want to divide the numerator and denominator by $e^x$ (i.e., the factor driving both the numerator and denominator to infinity). This reveals a limiting value of $5$:

$$\lim_{x \rightarrow \infty} \frac{10e^x - 21}{2e^x} = \lim_{x \rightarrow \infty} \frac{10 - \frac{21}{e^x}}{2} = \lim_{x \rightarrow \infty} \frac{10 - 0}{2} = 5$$

On the right, it is helpful to write things first with a single radical and then divide the numerator and denominator of the resulting fraction inside that radical by the highest power of $x$ present in the denominator (i.e., $x$):

$$\lim_{x \rightarrow \infty} \frac{5\sqrt{x}}{\sqrt{x-1}} = \lim_{x \rightarrow \infty} 5 \sqrt{\frac{x}{x-1}} = \lim_{x \rightarrow \infty} 5 \sqrt{\frac{1}{1-\frac{1}{x}}} = \lim_{x \rightarrow \infty} 5 \sqrt{\frac{1}{1-0}} = 5$$

As $f(x)$ is bound (for $x \gt 1$) between two functions with the same limiting value (namely, $5$) as $x \rightarrow \infty$, the Squeeze Theorem does indeed apply, and we have

$$\displaystyle{\lim_{x \rightarrow \infty} f(x) = 5}$$
15. Evaluate $\displaystyle{\lim_{x \rightarrow -\infty} \frac{\sin x}{x^3}}$; provide a graphical interpretation for the limit.

Note that as $x \rightarrow -\infty$, we at least have $-1 \lt \sin x \lt 1$. This suggests we might be able to use the Squeeze Theorem, as

$$\frac{-1}{x^3} \lt \frac{\sin x}{x^3} \lt \frac{1}{x^3}$$

Given the limit of the left and right sides as $x \rightarrow -\infty$ are both $0$, the Squeeze Theorem does indeed apply and the limit we seek must be $0$ as well.

Graphically, we have a horizontal asymptote (on the left) of $y = 0$.

16. A tank contains $5000$ L of pure water. Brine that contains $30$ g of salt per liter of water is pumped into the tank at a rate of $25$ L/min. Find an expression for the concentration of salt after $t$ minutes (in grams per liter). What happens to the concentration as $t \rightarrow \infty$? How does this answer make intuitive sense?

Concentration is mass over volume.

At time $t$, the mass of salt in the solution will be given by $(25)(30)(t)$ (since there are $t$ minutes, each minute 25 liters of brine are added to the solution, and each liter contains 30 grams of salt).

The total volume of the solution after $t$ minutes is $5000 + 25t$.

Thus, the overall concentration is given by

$$C(t) = \frac{(25)(30t)}{5000 + 25t} = \frac{30t}{200+t}$$

As $t \rightarrow \infty$, $C(t) \rightarrow 30$.

This is reasonable, since as $t$ gets larger and larger, the amount of water that was originally in the tank becomes a fraction of the total volume and can be disregarded as $t \rightarrow \infty$.